Problem: Find the equation of the plane containing the points $(-3,4,-2),$ $(1,4,0),$ and $(3,2,-1).$  Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$
Let $\mathbf{a} = \begin{pmatrix} -3 \\ 4 \\ -2 \end{pmatrix},$ $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \\ 0 \end{pmatrix},$ and $\mathbf{c} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}.$  Then the normal vector of the plane is orthogonal to both
\[\mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}\]and
\[\mathbf{c} - \mathbf{a} = \begin{pmatrix} 6 \\ -2 \\ 1 \end{pmatrix}.\]So to compute the normal vector, we take the cross product of these vectors:
\[\begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} 6 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ -8 \end{pmatrix}.\]We can scale this vector, and take $\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ as the normal vector.  Then the equation of the plane is of the form
\[x + 2y - 2z + D = 0.\]Substituting the coordinates of any of the points, we find that the equation of the plane is $\boxed{x + 2y - 2z - 9 = 0}.$